Question about Hatcher's book CW complex I am currently reading in Hatcher's book at page 522 about the construction of open sets in a CW complex. They start with an arbitrary set $A \subset X$ and want to construct an open neighborhood $N_{\varepsilon}(A)$.
You might also look at page 4 in this reference top of page 4
The construction is inductive over the skeleta $X^n$. Hence, he constructs a sequence of $N_{\varepsilon}^n(A)$ sets, where $N_{\varepsilon}^n(A)$ is an open neighbourhood of $A \cap X^n$.
He starts by saying: Take $N_{\varepsilon}^0 = A \cap X^0$. I don't see why this should be open. This is an arbitrary set of discrete points, why should this be open?
I also don't see why this induction works: I mean, there does not need to be an intersection between $A$ and a particular $n+1$ cell, so how do you construct this open $\varepsilon_{\alpha}$ neighborhood. The only possible explanation to me is: If the intersection is trivial, then you are allowed to take any epsilon ball and otherwise you take an open covering of $A$ where each ball has a distance less than $\epsilon_{\alpha}$ away from $A$, correct?
And maybe you could comment on this spherical coordinates thing. What is a Cartesian product with respect to spherical coordinates? Does this mean that you the radius is taken between $(1-\varepsilon_{alpha},1]$ and the angle such that the second product factor is completely covered by all combination $(r,\theta)$. Maybe anybody here could add a few more details on this?
 A: Spherical coordiantes are a way to express a point on the $n$-sphere via angular coordinates 
$$\phi, \theta_1,...,\theta_{n-1} \qquad \text{ where }\; 0\le\phi\le2\pi,\;\; 0\le\theta_i\le\pi$$
Then if you want to define in point in the $n$-ball, you have to add a radius $0\le r\le 1$. The topology on $D^n$ goes well with this description, in particular, there is a quotient map $[0,1]\times[0,2π]×[0,π]×...×[0,π]\to D^n$
$$
(r,\phi,\theta_1,\dots,\theta_{n-1}) \mapsto 
\pmatrix{
r \sin\theta_1 \dots \sin\theta_{n-1} \sin\phi \\
r \sin\theta_1 \dots \sin\theta_{n-1} \cos\phi \\
r \sin\theta_1 \dots \cos\theta_{n-1} \\
\vdots \\
r \sin\theta_1 \sin\theta_2 \cos\theta_3 \\
r \sin\theta_1 \cos\theta_2 \\
r \cos \theta_1 \\ }
$$
so if your set $N^{n-1}_ε(A)$ is already open in the sphere $\partial D^n_\alpha$, then adding an open range of $r\in(1-ε_\alpha,\ 1]$ gives an open set in $D_\alpha^n$. Also see this wiki article.
In dimension zero any set in $X^0$ is open (and closed) as $X^0$ has the discrete topology.
If $\Phi^{-1}_\alpha(N^{n-1}_ε(A))$ is empty in $D^n_\alpha$, then you could simply take the empty set in that ball.
